$g(x)=(3x^5)(6x^4+x)^4$ Find $g'(x)$. Choose 1 answer: Choose 1 answer: (Choice A) A $1944x^{23}(6x^3+1)^3(9x^3+1)$ (Choice B) B $6x^9(39x^3+5)$ (Choice C) C $189x^8$ (Choice D) D $(378x^8+27x^5)(6x^4+x)^3$
Explanation: $(3x^5)(6x^4+x)^4$ is a product of a composite function and another function. Let... $u(x)=3x^5$ $v(x)=x^4$ $w(x)=6x^4+x$... then $g(x)=u(x)\cdot v\Bigl(w(x)\Bigr)$. To find $g'(x)$, we will need to use the product rule and the chain rule! $\begin{aligned} &\phantom{=}\dfrac{d}{dx}\left[u(x)\cdot v\Bigl(w(x)\Bigr)\right] \\\\ &=u'(x)\cdot v\Bigl(w(x)\Bigr)+u(x)\cdot\dfrac{d}{dx}\left[v\Bigl(w(x)\Bigr)\right] \gray{\text{Product rule}} \\\\ &=u'(x)\cdot v\Bigl(w(x)\Bigr)+u(x)\cdot v'\Bigl(w(x)\Bigr)\cdot w'(x) \gray{\text{Chain rule}} \end{aligned}$ Let's differentiate $u$, $v$, and $w$ : $u'(x)=15x^4$ $v'(x)=4x^3$ $w'(x)=24x^3+1$ Now we can plug the equations for $u$, $v$, $w$, $u'$, $v'$, AND $w'$ into the expression we got: $\begin{aligned} &\phantom{=}u'(x)\cdot{ v\Bigl(w(x)\Bigr)}+u(x)\cdot{ v'\Bigl(w(x)\Bigr)}\cdot w'(x) \\\\ &=15x^4\cdot{(6x^4+x)^4}+3x^5\cdot{4(6x^4+x)^3}\cdot (24x^3+1) \\\\ &=(15x^4\cdot (6x^4+x) + 288x^8+12x^5)\cdot (6x^4+x)^3 \\\\ &=(378x^8+27x^5)(6x^4+x)^3 \end{aligned}$ In conclusion: $g'(x)=(378x^8+27x^5)(6x^4+x)^3$